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Question Number 122732 by fajri last updated on 19/Nov/20
find critical point from    Differential Equation System    (dy/dx) = x − y   (dy/dx) = x^2  +y^2  − 2
$${find}\:{critical}\:{point}\:{from}\: \\ $$$$\:{Differential}\:{Equation}\:{System} \\ $$$$ \\ $$$$\frac{{dy}}{{dx}}\:=\:{x}\:−\:{y}\: \\ $$$$\frac{{dy}}{{dx}}\:=\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:−\:\mathrm{2}\: \\ $$
Answered by bemath last updated on 19/Nov/20
(1) let y = x−u ⇒ (dy/dx) = 1−(du/dx)  ⇒ 1−(du/dx) = x−(x−u)  ⇒ 1−u = (du/dx) ; (du/(1−u)) = dx   ⇒ln ∣1−u∣ = x + c   ⇒ 1−u = ±Ke^x  ; 1−(x−y)=±Ke^x   ⇒ y = ±Ke^x +x−1  ⇒y′ = ±Ke^x +1 = 0 ;   ⇒ e^x  = (1/K) , where (1/K)>0  ⇒ x = −ln K and y = −Ke^(−ln K) −ln K+1   y = −K((1/K))−ln K +1 = −ln K   Hence critical point of y=f(x)=−Ke^x +x−1  is ( −ln K, −ln K ).
$$\left(\mathrm{1}\right)\:{let}\:{y}\:=\:{x}−{u}\:\Rightarrow\:\frac{{dy}}{{dx}}\:=\:\mathrm{1}−\frac{{du}}{{dx}} \\ $$$$\Rightarrow\:\mathrm{1}−\frac{{du}}{{dx}}\:=\:{x}−\left({x}−{u}\right) \\ $$$$\Rightarrow\:\mathrm{1}−{u}\:=\:\frac{{du}}{{dx}}\:;\:\frac{{du}}{\mathrm{1}−{u}}\:=\:{dx} \\ $$$$\:\Rightarrow\mathrm{ln}\:\mid\mathrm{1}−{u}\mid\:=\:{x}\:+\:{c}\: \\ $$$$\Rightarrow\:\mathrm{1}−{u}\:=\:\pm{Ke}^{{x}} \:;\:\mathrm{1}−\left({x}−{y}\right)=\pm{Ke}^{{x}} \\ $$$$\Rightarrow\:{y}\:=\:\pm{Ke}^{{x}} +{x}−\mathrm{1} \\ $$$$\Rightarrow{y}'\:=\:\pm{Ke}^{{x}} +\mathrm{1}\:=\:\mathrm{0}\:;\: \\ $$$$\Rightarrow\:{e}^{{x}} \:=\:\frac{\mathrm{1}}{{K}}\:,\:{where}\:\frac{\mathrm{1}}{{K}}>\mathrm{0} \\ $$$$\Rightarrow\:{x}\:=\:−\mathrm{ln}\:{K}\:{and}\:{y}\:=\:−{Ke}^{−\mathrm{ln}\:{K}} −\mathrm{ln}\:{K}+\mathrm{1}\: \\ $$$${y}\:=\:−{K}\left(\frac{\mathrm{1}}{{K}}\right)−\mathrm{ln}\:{K}\:+\mathrm{1}\:=\:−\mathrm{ln}\:{K}\: \\ $$$${Hence}\:{critical}\:{point}\:{of}\:{y}={f}\left({x}\right)=−{Ke}^{{x}} +{x}−\mathrm{1} \\ $$$${is}\:\left(\:−\mathrm{ln}\:{K},\:−\mathrm{ln}\:{K}\:\right). \\ $$$$\: \\ $$
Commented by fajri last updated on 19/Nov/20
thank Sir, I Like it :)
$$\left.{thank}\:{Sir},\:{I}\:{Like}\:{it}\::\right) \\ $$

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